00:01
In this question, we are asked to find the sum of the given series.
00:04
To do that, we will first write this series as negative series 1 over, sorry, not 1 over n, negative 1 third to the n, divided by n, n from 1 to infinity.
00:18
What we did, we just moved negative 1 3 to the n to the numerator, that's all.
00:23
Now, we'll start with a series 1, with a series for 1 over 1 minus x, which is a geometric series and equals to the series x to the n, n from 0 to infinity.
00:36
Then, what we are going to integrate this series.
00:40
Recall that the interval of 1 over 1 minus x equals to negative ln of the absolute value of 1 minus x.
00:49
On the other hand, if we integrate the series on the right -hand side, we are going to get the series x to the n plus 1, divide by n plus 1, plus the constant of integration c where n is running from 0 to infinity.
01:09
Now, to find the constant of iteration c, we will plug in x equals 0 in both sides of the equation.
01:16
Negative ln1 equals to, if we plug in x equal 0 in the series here, you will get that every term equals to 0.
01:23
And we are going to get c equals to negative ln1...